Question

$\text{Standard enthalpy of vaporisation for CCl}_4 \text{ is 30.5 kJ mol}^{-1}. \text{ Heat required for vaporisation of } 284 \, \text{g of CCl}_4 \text{ at constant temperature is } \_\_\_\_ \, \text{kJ.} \\ \text{(Given molar mass in g mol}^{-1}\text{; C = 12, Cl = 35.5)}$

Answer 1

Correct Answer: 56

\(\Delta H_{\text{vap}}^\circ \, \text{for } \, \text{CCl}_4 = 30.5 \, \text{kJ/mol}\)

Mass of \( \text{CCl}_4 = 284 \, \text{g} \)  
Molar mass of \( \text{CCl}_4 \):  
\(= 154 \, \text{g/mol}\)

Moles of \( \text{CCl}_4 \):  
\(= \frac{284}{154} = 1.844 \, \text{mol}\)

\(\Delta H_{\text{vap}}^\circ \, \text{for 1 mole} = 30.5 \, \text{kJ/mol}\)

\(\Delta H_{\text{vap}}^\circ \, \text{for 1.844 moles} = 30.5 \times 1.844\)

\(= 56.242 \, \text{kJ}\)

The Correct Answer is:= \(56.242 \, \text{kJ}\)